Sign of the Product of an Array

easy math scan

Problem

Return 1 if the product of nums is positive, −1 if negative, 0 if any element is 0.

Inputnums = [-1,-2,-3,-4,3,2,1]

Output1

Four negatives → product positive.

nums

def array_sign(nums):
    sign = 1
    for x in nums:
        if x == 0: return 0
        if x < 0: sign = -sign
    return sign

function arraySign(nums) {
  let sign = 1;
  for (const x of nums) {
    if (x === 0) return 0;
    if (x < 0) sign = -sign;
  }
  return sign;
}

class Solution {
    public int arraySign(int[] nums) {
        int sign = 1;
        for (int x : nums) {
            if (x == 0) return 0;
            if (x < 0) sign = -sign;
        }
        return sign;
    }
}

int arraySign(vector<int>& nums) {
    int sign = 1;
    for (int x : nums) {
        if (x == 0) return 0;
        if (x < 0) sign = -sign;
    }
    return sign;
}

Explanation

We never actually multiply the numbers — that could overflow and we don't care about the value, only the sign. So we just track whether the running sign is positive or negative.

Start with sign = 1. Walk through each element: if it's 0, the whole product is 0, so we can short-circuit and return 0 immediately. If it's negative, we flip the sign with sign = -sign. Positive numbers don't change anything.

At the end, sign is +1 if there were an even number of negatives and -1 if odd — exactly the sign of the real product.

Example: nums = [-1, -2, -3, -4, 3, 2, 1] has four negatives. Each one flips the sign, and four flips land back on +1, so the answer is 1.

It works because a product's sign depends only on how many negative factors there are (and whether any factor is zero), not on the magnitudes.

Time: O(n) Space: O(1)